1.

A particle moves along a straight line accordingto the law s=t^3-3t^2+5t. Find its velocity and acceleration at the end of 1 sec.

Answer»

SOLUTION :`s=t^3-3t^2+5trArr(ds)/(dt)=3t^2-6t+5rArr(d^2s)/(dt^2)=6t-6` At the end of 1 SEC The velocity `=(ds)/(dt)]_(r=1)=2` The ACCELERATION `=(d^2s)/(dt^2)]_(t=1)=0`


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